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# M15#11

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07 May 2012, 23:21
Is $$|x - y| \gt |x + y|$$ ?

1) $$x^2 - y^2 = 9$$
2) $$x - y = 2$$

From S1, can we assume that x has to equal 5 and y has to equal 4 so when you square x and y, their difference would be 9. Thus S1 would be sufficient.

S2 is insufficient.

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07 May 2012, 23:30
Samwong wrote:
Is $$|x - y| \gt |x + y|$$ ?

1) $$x^2 - y^2 = 9$$
2) $$x - y = 2$$

From S1, can we assume that x has to equal 5 and y has to equal 4 so when you square x and y, their difference would be 9. Thus S1 would be sufficient.

S2 is insufficient.

Why would you assume that? Why not x=-5 and y=-4? Or x=3 and y=0? $$x^2 - y^2 = 9$$ has infinitely many solutions for x and y, and only one of them is x=5 and y=4.

Is $$|x - y| \gt |x + y|$$ ?

(1) $$x^2 - y^2 = 9$$ --> $$(x-y)(x+y)=9$$. Not sufficient.
(2) $$x - y = 2$$ --> don't know the value of $$x+y$$. Not sufficient.

(1)+(2) Since from (2) $$x - y = 2$$ then from (1) $$(x-y)(x+y)=2(x+y)=9$$ --> $$x+y=4.5$$. Sufficient.

Hope it helps.
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07 May 2012, 23:54
Thanks Bunuel for the reply. That make sense now. I did a similar problem on MGMAT CAT. The difference is that on the MGMAT problem there is a constrain of x < y < 0

If x and y are integers such that x < y < 0, what is x – y?

(1) (x + y)(x – y) = 5

(2) xy = 6

Official Explanation:
The problem stem tells you that x and y are both negative integers, and x is less than y. You are asked for the value of x – y. Note that in order to answer this question, you might not need the separate values of x and y; however, if you can find those separate values, then you can solve for x – y or any other “combo.”

(1) SUFFICIENT: This statement is tricky. One approach is to distribute the left side to get the difference of squares:

x2 – y2 = 5

Now, since both x and y are integers, x2 and y2 are integers as well. Let’s look at the sequence of squares, and consider their differences (we can leave out zero, since neither x nor y can be zero):

1 4 9 16 25 36…
Diff = 3 5 7 9 11…

As you can see, the difference between squares grows as the squares themselves get larger. The only difference between two squares that equals 5 is the difference between 4 and 9. Since x and y are both negative, this tells us that x = –3 and y = –2; therefore, x – y = –1.

Alternatively, we could note that both x + y and x – y are themselves integers. Looking at the statement, we have
(x + y)(x – y) = 5
int × int = 5

The only possible integer pairs that give 5 as a product are (5, 1) and (-5, -1), since 5 is prime. Now, because both x and y are negative, the (5, 1) pair won’t work either way (either with x + y = 5 or with x + y = 1). So let’s try (-5, -1):

x + y = –5
x – y = –1

Adding these two equations, we get 2x = -6, or x = -3. Substituting back into x + y = –5, we get y = –2. (If we had assigned x + y = –1 and x – y = –5, we would have gotten y = 2, which doesn’t fit the problem constraints.)

(2) INSUFFICIENT: There are two pairs of integers that fit the constraint x < y < 0 and the statement xy = 6: (-3)(-2) = 6 AND (-6)(-1) = 6.

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08 May 2012, 00:02
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Samwong wrote:
Thanks Bunuel for the reply. That make sense now. I did a similar problem on MGMAT CAT. The difference is that on the MGMAT problem there is a constrain of x < y < 0

If x and y are integers such that x < y < 0, what is x – y?

(1) (x + y)(x – y) = 5

(2) xy = 6

Actually you have more than one constraint:
(i) x and y are integers;
(ii) x<y<0

If x and y are integers such that x<y<0 what is x-y?

(1) (x+y)(x-y)=5. x and y are integers means that both x+y and x-y are integers. So, we have that the product of two integer factors equal to 5. There are only two combination of such factors possible: (1, 5) and (-1, -5). Since given that x and y are both negative then the first case is out, so x-y is either -1 or -5, but it can not be -5, because in this case x+y must be -1 and no sum of two negative integers yields -1. Hence x-y=-1. Sufficient.

(2) xy= 6. If x=-3 and y=-2 then x-y=-1 but if x=-6 and y=-1 then x-y=-5. Not sufficient.

Hope it helps.
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10 May 2012, 03:40
Just for information, if a questions as such if we square on both sides, then it does not change anything..am i correct in stating this? for examle |x-y| > |x+y|...then (x-y)2 = (x=y)2?
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10 May 2012, 03:56
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pavanpuneet wrote:
Just for information, if a questions as such if we square on both sides, then it does not change anything..am i correct in stating this? for examle |x-y| > |x+y|...then (x-y)2 = (x=y)2?

Yes, since both sides of the inequality are non-negative then we can square it and write: (x-y)^2>(x+y)^2.

Generally:
A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

Hope it helps.
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29 Nov 2012, 22:41
For the life of me I can't see a situation where abs (x+y) is less than 2, which would make statement B sufficient.
Re: M15#11   [#permalink] 29 Nov 2012, 22:41
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# M15#11

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